Two-or-three-dimensional-motion, Mechanical Engineering

Net force acting on particle of mass 2 kg is 10N in north direction.At t=o, particle was moving eastwards with 10m/s. Find displacement and velocity of particle after 2 seconds.

given force =F=10N

mass M=2kg

velocity u=10m/s

then acceleration a=F/M=10/2=5m/s^2

then displacement=s=ut+1/2at^2

here u=initial velocity



then we get the displacement is=30m


then we get velocity as 20m/s.

Posted Date: 3/8/2013 7:08:50 AM | Location : United States

Initial velocity is 10 m/s in x-direction.

Now a force of 10N acts in y-direction 

So , using F = ma 

               a = f/m = 10/2 = 5m/s2in y-direction 

After 2 seconds velocity in y-direction would become 10 m/s

Now we can take the resultant of the two velocities i.e 10 m/s in x-direction & 10m/s in y-direction

Therefore Resultant velocity = (102+102)1/2

                    = (200)1/2 and 450 North of East

Displacement = velocity x time 

                    = (200)1/2x 2

                    = 20x(2)1/2m 450 north of east

Posted by | Posted Date: 3/8/2013 7:09:55 AM

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